Angular Acceleration Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. The angular acceleration is also known as

1. Moving acceleration
2. Rotational acceleration
3. Angle acceleration
4. Circular acceleration

Ans. The correct answer is b. Rotational acceleration

Explanation: Rotational acceleration is another name for angular acceleration. It is a mathematical representation of the variation in angular velocity over time.

Ques. What is the formula to find the angular acceleration?

1. ɑ = dω/dt
2. ɑ = dω x dt
3. ɑ = dω/dt²
4. ɑ = dω + dt

Ans. The correct answer is a. ɑ = dω/dt

Explanation: The time rate of change of angular velocity (ω) is known as angular acceleration. The formula of angular acceleration is given by ɑ = dω/dt.

Ques. Angular acceleration is a

1. Dimensionless quantity
2. Scalar quantity
3. Vector quantity
4. None of the above

Ans. The correct answer is c. Vector quantity

Explanation: Angular acceleration is a vector quantity and its direction is perpendicular to the plane in which the rotation takes place.

Ques. The angular acceleration is represented as

1. ɑ
2. β
3. ℽ
4. τ

Ans. The correct answer is a. ɑ

Explanation: The angular acceleration is represented by the symbol alpha (ɑ).

Ques. When the angular velocity is constant, the angular acceleration is

1. -1
2. 1
3. 0
4. None of the above

Ans. The correct answer is c. 0

Explanation: Constant angular velocity represents there is no change in angular velocity. Therefore angular acceleration is 0.

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Short Answers Questions [2 Marks Questions]

Ques. What is angular acceleration? What is the SI unit of angular acceleration?

Ans. The rate at which angular velocity changes over time is known as angular acceleration, also known as rotational acceleration.

It is denoted by the symbol alpha (ɑ) and its SI unit is radian per square second (rad/s²).

Ques. Can the angular acceleration be zero?

Ans. The rate of change of angular velocity gives angular acceleration. If the change in angular velocity is constant, then angular acceleration becomes zero.

Ques. What is angular velocity?

Ans. A pseudovector that represents how an object's angular position or orientation varies over time, or how quickly an object rotates (spins or circles) around an axis of rotation and how quickly the axis itself changes direction, is known as an angular velocity or angular frequency vector.

Ques. List three examples of angular acceleration.

Ans. The various examples of angular acceleration are

• Car Turning a Corner
• Pendulum Clock
• Spinning Top
• Merry-Go-Round
• Diver Performing a Somersault
• Figure Skater Spinning

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Long Answers Questions [3 Marks Questions]

Ques. When a ceiling fan is switched on, it makes 10 revolutions in the first 3 seconds. Assuming a uniform angular acceleration, how many rotations it will make in the next 3 seconds?

Ans. Angle turned in first 3 seconds, θ₃ = 2π x 10 = 20π

Using the equation of motion

θ₃ = ω₀t + (1/2)ɑt²

⇒ 20π = 0 + (1/2)ɑ x 33

⇒ ɑ = 40π/9 rad/s²

Now the angle turned in 6 sec from the starting

θ₆ = ω₀t + (1/2)ɑt²

⇒ θ₆ = 0 + (1/2)(40π/9) x 6² = 80π rad

Angle turned between t = 3s to t = 6s

θ = θ₆ - θ₃ = 80π - 20π = 60π

Number of revolution = 60π/2π = 30 revolutions

Ques. A wheel has an angular acceleration of 3 rad/s² and an initial angular speed of 2 rad/s. How much it will rotate in 2 seconds?

Ans. Given

• Angular acceleration, a = 3 rad/s²
• Initial angular speed, ωi = 2 rad/s
• Time, t = 2 s

Using the equation of rotational motion, the displacement of the wheel in time t is given by

θ = ω_it + (1/2)at²

On substituting the values, we get

θ = (2 x 2) + (1/2) x 3 x 2²

⇒ θ = 4 + 6 = 10 rad

Ques. A car is moving at a speed of 54 km/hr. The diameter of its wheel is 0.6 m. If the wheels are stopped in 10 rotations by applying brakes, then what will be the angular retardation produced by the brakes?

Ans. Given

• Speed of the car, v = 54 km/h = 54 x 5/18 = 15 m/s
• The radius of the wheel, r = 0.6 /2 = 0.3 m
• Angular distance covered by the car, s = 10 rotations

Circumference of one rotation = 2πr = 2 x 3.14 x 0.3 = 1.88 m

Therefore, the total distance covered in 10 rotations = 10 x 1.88 = 18.8 m

Also, the initial angular speed, ω_i = v/r = 15/0.3 = 50 rad/s

Using the equation of rotational motion, we have

 ω²_f = ω²_i + 2as

⇒ Angular acceleration, a = ( ω²f - ω²i)/2s

⇒ a = (0 - 50²) / (2 x 18.8) = - 66.49 rad/s²

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Very Long Answers Questions [5 Marks Questions]

Ques. The wheel of a car is rotating at the rate of 1200 revolutions per minute. On pressing the accelerator for 10 seconds, it starts rotating at 4500 revolutions per minute. Calculate the angular acceleration of the wheel.

Ans. Initially, the frequency of the revolution, f_i = 1200 revolutions per minute = 1200 / 60 revolutions per second

⇒ the frequency of the revolution, f_i = 20 revolutions per second

Therefore initial angular velocity of the wheel is given by

ω_i = 2πf_i

⇒ ω_i = 2 x 3.14 x 20 = 125.6 rad/s

Finally, the frequency of the revolution, f_f = 4500 revolutions per minutes = 4500 / 60 revolutions per second

⇒ the frequency of the revolution, f_f = 75 revolutions per second

Therefore final angular velocity of the wheel is given by

ω_f = 2πf_f

⇒ ω_f = 2 x 3.14 x 75 = 471 rad/s

Angular acceleration is given by

a = (ω_f - ω_i) / t

⇒ a = (471 - 125.6) / 10 = 34.54 rad/s²

Ques. The moment of inertia of a body about a given axis is 2.4 kg m². What will be the time required to produce rotational kinetic energy of 750 J, if the angular acceleration of 5 rad/s² is applied about that axis?

Ans. Given

• Moment of inertia of the body, I = 2.4 kg m²
• The kinetic energy of the body, E = 750 J
• Angular acceleration, a = 5 rad/s²

Initially, the body is at rest, therefore its initial angular velocity, ω_i = 0

Now, the kinetic energy of a rotating body is given by

E = (1/2)Iω²

⇒ 750 = (1/2) x 2.4 x ω_f²

Where ωf is the final angular velocity of the body.

⇒ ω_f² = (750 x 2) / 2.4 = 625

⇒ ω_f = 25 rad/s

Now, angular acceleration is given by

a = (ω_f – ω_i) / t

⇒ t = (ω_f – ω_i) / a = (25 - 0) / 5

⇒ t = 5 seconds

Ques. A cord is wound over the rim of a flywheel of mass 20 kg and radius 25 cm. A mass of 2.5 kg is attached to the cord and is allowed to fall under gravity. Calculate the angular acceleration of the flywheel.

Ans. Given

• The mass of the flywheel, m = 20 kg
• The radius of the flywheel, r = 25 cm = 0.25 m

The moment of inertia of the flywheel about its axis is given by

I = mr²/2

⇒ I = (20 x 0.25²) / 2 = 625 x 10^-3 kg m²

Now, when a mass of 2.5 kg is attached to the cord, then a downward force equal to the weight of the mass attached is acting the flywheel. Therefore force acting on the flywheel is given by

F = mg = 2.5 x 10 = 25 N

Torque acting on the flywheel

τ = F x r = 25 x 0.25 = 625 x 10^-2 N m

Angular acceleration of the flywheel is given by

a = τ/I = (625 x 10^-2) / (625 x 10^-3)

⇒ a = 10 rad s²

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